a small ball of mass .85 kg is attached to one end of a 1 m long massless rod, and the other end of the rod is hung from a pivot. When the resulting pendulum is 25 degrees from the vertical, what is the magnitude of the torque about the pivot?
AP physics - Christiaan, Monday, January 28, 2008 at 7:09am
Torque is calculated by the formula:
t=r.F.sin(a) where a is the angle between the rod and the force acting on the rod(=25°),
r is the distance from the force to the pivot point(which is 1m),
and F is the gravity force acting on the ball(=m.g = 0.85kg.10m/s² =8.5 N)
So, the torque equals:
t=1m . 8,5N . sin(25°) = 3,59 Nm